[Interest] How to create QObject-derived class instance from class name

[Sina Dogru]

I thank you for your answers and solutions.

Before I wrote, I tried to use QMetaType::create function to instantiate an
instance but since QObject have deleted copy constructor and copy
assignment operator, I have ended up using QObject pointer with QMetaType..
And when I see that QMetaType::create() function, it just create a pointer
to QObject but not the QObject instance.. So I thought QMetaType is not for
the QObject-derived classes..

But I have not realised that QMetaType::metaObjectForType function returns
QMetaObject instance for the class not for the pointer to class.

For now there is a design choice which I feel weak myself, to do factory or
to use QMetaType for creating instances.

Thank you,
Sina

2016-03-01 18:51 GMT+02:00 André Somers <andre at familiesomers.nl>:

>
>
> Op 01/03/2016 om 17:21 schreef Thiago Macieira:
>
>> On terça-feira, 1 de março de 2016 17:06:49 PST André Somers wrote:
>>
>>> The meta *object* system has no registration.
>>>>
>>>> The meta *type* system requires that the registered type be default-
>>>> constructible and copyable, but QObject is not copyable. Therefore,
>>>> QObject- derived classes cannot be registered with the meta type system.
>>>>
>>> Am I completely misinterpretting the documentation then?
>>> http://doc.qt.io/qt-5/qmetatype.html#metaObject
>>>
>>> If I read that correctly, you can register a
>>> pointer-to-a-QObject-derived-class-instance and use that. So, indeed,
>>> you do not register the type, but the type*. And that has no problems
>>> with being default constructed or copied.
>>>
>> Correct. You can't register a QObject class with the meta type system,
>> but you
>> can register a pointer to a QObject class. The problem is that
>> QMetaType::create() will then create a pointer, not the object.
>>
> Of course. But... Again, if I read it correctly, you *can* then get the
> QMetaObject from QMetaType, and using that, you can create an actual
> instance.
>
> Just tested this trivial example:
>
> //main.cpp
> int main(int argc, char *argv[])
> {
>     qRegisterMetaType<TestClass*>(); //this could be elsewhere of course
>
>     QCoreApplication a(argc, argv);
>
>     auto tId = QMetaType::type("TestClass*"); //just using the class name
> with an *
>     auto metaObject = QMetaType::metaObjectForType(tId);
>     QObject* instance = metaObject->newInstance();
>
>     return a.exec();
>     delete instance;
> }
>
> //testclass.h
> class TestClass: public QObject
> {
>     Q_OBJECT
>
> public:
>     Q_INVOKABLE TestClass();
> };
>
> Q_DECLARE_METATYPE(TestClass*)
>
> //testclass.cpp
> TestClass::TestClass()
> {
>     qDebug() << "TestClass instance created";
> }
>
> Which prints out "TestClass instance created" on the console. My
> conclusion is that it works, and that you _can_ create a QObject derived
> instance with just the class name.
>
> Again: I am not claiming that abusing this to skip defining a factory is a
> good idea. Just that it is possible.
>
>
> André
>
>
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